Applications of Integral Calculus to Root Solving

In summary, the conversation discusses alternative methods for finding the roots of quadratics and the implications of calculus in algebra. The area under a curve can be represented as an integral and can be maximized when the total area between the two roots is found. However, this method is circular and ultimately leads back to the original equation.
  • #1
lapo3399
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As a Grade 12 student that is often required to find the roots of quadratics for math, physics, and chemistry problems, I wondered whether there would be any methods for solving these problems excepting the quadratic formula. I was pondering the implications of calculus in algebra and, although this may seem much more complicated than the quadratic formula itself, have determined something interesting regarding roots and integrals.

If the area under a curve is found for a function f(x)

[tex] A = \int_{a}^{b}\ f(x) dx [/tex]

which may also be represented as

[tex] A(x) = \int_{a}^{a+c}\ f(x) dx [/tex]

where h is the different between a and b, then

[tex] A(x) = F(a+c) - F(a) [/tex]

As A(x) will be maximized when the total area between the two roots is found (assuming no improper integrals or infinite areas), then

[tex] a(x) = f(a+c) - f(a)[/tex]
[tex] 0 = f(a+c) - f(a)[/tex]
[tex] f(a) = f(a+c)[/tex]

It is rather obvious that this means that the function has equal values (0 as the two x values lie on the x-axis) at f(a) or the first root and f(a+c) or the second root, but I must ask something that has been puzzling me concerning this rather meaningless conclusion : if the fact that the area function is maximized causes f(a) to equal f(a+c), and considering the fact that, for example, a quadratic has an infinite number of solutions for f(a) = f(a+c) that are not restricted to the roots of the equation, why should the maximization condition be necessary to produce the equation f(a) = f(a+c)?

The best explanation that I have for this is that, assuming c is remaining constant, the rate of change in area on the left will be the negation of the rate of change on the right, and so there is no maximum for the area function AS I have defined it. That is, if I were to take a quadratic and decrease a, the area concerned would increase/decrease by a certain amount, but the change in a+c would compensate for this with an equal area change on the right, thus keeping the area constant. The only way I see of defining this better is defining c as a non-constant, as it obviously will change depending on the function.

Please provide any insight that you have!

Thanks,
lapo3399
 
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  • #2
lapo3399 said:
If the area under a curve is found for a function f(x)

[tex] A = \int_{a}^{b}\ f(x) dx [/tex]

which may also be represented as

[tex] A(x) = \int_{a}^{a+c}\ f(x) dx [/tex]

where h is the different between a and b, then

[tex] A(x) = F(a+c) - F(a) [/tex]

A does not depend on x, but rather on a and c. I'll assume a is constant and c is what you're after.

lapo3399 said:
As A(x) will be maximized when the total area between the two roots is found (assuming no improper integrals or infinite areas), then

[tex] a(x) = f(a+c) - f(a)[/tex]
[tex] 0 = f(a+c) - f(a)[/tex]
[tex] f(a) = f(a+c)[/tex]

Lost me there. I'll assume a is the derivative of A with respect to c (still not sure where x comes in here). But then F(a) is a constant, and so we actually have

[tex] a(c) = f(a+c)[/tex]
[tex] 0 = f(a+c)[/tex]

which is practically a tautology.
 
  • #3
Hi Lapo. One problem of course is that your method is fundamentally a circular argument (integrating the function and then setting the derivative to zero). Had you made no errors in the process then your "solution" could only have been to arrive back at the starting point, that is with the original equation f(x)=0

[tex] A(x) = F(a+c) - F(a) [/tex]
There's a problem right there as this is really a function of both "a" and "c". If you wanted to proceed like this you should have written,

[tex] A(x1,x2) = F(x2) - F(x1) [/tex].

You then would have taken the two partial derivatives and set them to zero, each one yeilding nothing but the original problem : f(x1)=0, f(x2)=0.

Easier would have been to simply write [tex] A(x) = F(x) + const [/tex] and then differentiate that to give f(x)=0.
 
Last edited:

1. How is integral calculus used to solve for roots?

Integral calculus is used to solve for roots by finding the area under a curve. This area represents the definite integral of the function, and when it equals zero, it indicates the root of the function. By using various integration techniques, the root can be approximated with a high degree of accuracy.

2. What is the benefit of using integral calculus to solve for roots?

The benefit of using integral calculus to solve for roots is that it provides a more precise and accurate solution compared to other methods, such as graphical or numerical approaches. Additionally, it can be applied to a wide range of functions, making it a versatile tool for root solving.

3. Can integral calculus be used to solve for complex roots?

Yes, integral calculus can be used to solve for complex roots. By using complex integration techniques, the area under a complex function can be calculated to find the root. However, the process may be more complex and time-consuming compared to solving for real roots.

4. Are there any limitations to using integral calculus for root solving?

One limitation of using integral calculus for root solving is that it may not always provide an exact solution. Depending on the complexity of the function, the root may need to be approximated using numerical methods. Additionally, the process may become more challenging for higher-order polynomials or transcendental functions.

5. How is integral calculus applied in real-life situations for root solving?

Integral calculus is commonly used in engineering, physics, and economics to solve for roots in real-life situations. For example, it can be used to find the optimal solution for a production process, determine the trajectory of a projectile, or calculate the net present value of an investment. It is a powerful tool for solving real-world problems that involve finding roots.

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